Question:
The n candidates for a job have been ranked 1, 2, 3,..., n. Let x = rank of a randomly selected candidate, so that x has pmf:
![p(x) = \left \{ {{\frac{1}{n}\ \ x=1,2,3...,n} \atop {0\ \ \ Otherwise}} \right.](https://tex.z-dn.net/?f=p%28x%29%20%3D%20%5Cleft%20%5C%7B%20%7B%7B%5Cfrac%7B1%7D%7Bn%7D%5C%20%5C%20x%3D1%2C2%2C3...%2Cn%7D%20%20%5Catop%20%7B0%5C%20%5C%20%5C%20Otherwise%7D%7D%20%5Cright.)
(this is called the discrete uniform distribution).
Compute E(X) and V(X) using the shortcut formula.
[Hint: The sum of the first n positive integers is
, whereas the sum of their squares is
Answer:
![E(x) = \frac{n+1}{2}](https://tex.z-dn.net/?f=E%28x%29%20%3D%20%5Cfrac%7Bn%2B1%7D%7B2%7D)
or ![Var(x) = \frac{(n+1)(n-1)}{12}](https://tex.z-dn.net/?f=Var%28x%29%20%3D%20%5Cfrac%7B%28n%2B1%29%28n-1%29%7D%7B12%7D)
Step-by-step explanation:
Given
PMF
![p(x) = \left \{ {{\frac{1}{n}\ \ x=1,2,3...,n} \atop {0\ \ \ Otherwise}} \right.](https://tex.z-dn.net/?f=p%28x%29%20%3D%20%5Cleft%20%5C%7B%20%7B%7B%5Cfrac%7B1%7D%7Bn%7D%5C%20%5C%20x%3D1%2C2%2C3...%2Cn%7D%20%20%5Catop%20%7B0%5C%20%5C%20%5C%20Otherwise%7D%7D%20%5Cright.)
Required
Determine the E(x) and Var(x)
E(x) is calculated as:
![E(x) = \sum \limits^{n}_{i} \ x * p(x)](https://tex.z-dn.net/?f=E%28x%29%20%3D%20%5Csum%20%5Climits%5E%7Bn%7D_%7Bi%7D%20%5C%20x%20%2A%20p%28x%29)
This gives:
![E(x) = \sum \limits^{n}_{x=1} \ x * \frac{1}{n}](https://tex.z-dn.net/?f=E%28x%29%20%3D%20%5Csum%20%5Climits%5E%7Bn%7D_%7Bx%3D1%7D%20%5C%20x%20%2A%20%5Cfrac%7B1%7D%7Bn%7D)
![E(x) = \sum \limits^{n}_{x=1} \frac{x}{n}](https://tex.z-dn.net/?f=E%28x%29%20%3D%20%5Csum%20%5Climits%5E%7Bn%7D_%7Bx%3D1%7D%20%5Cfrac%7Bx%7D%7Bn%7D)
![E(x) = \frac{1}{n}\sum \limits^{n}_{x=1} x](https://tex.z-dn.net/?f=E%28x%29%20%3D%20%5Cfrac%7B1%7D%7Bn%7D%5Csum%20%5Climits%5E%7Bn%7D_%7Bx%3D1%7D%20x)
From the hint given:
![\sum \limits^{n}_{x=1} x =\frac{n(n +1)}{2}](https://tex.z-dn.net/?f=%5Csum%20%5Climits%5E%7Bn%7D_%7Bx%3D1%7D%20x%20%3D%5Cfrac%7Bn%28n%20%2B1%29%7D%7B2%7D)
So:
![E(x) = \frac{1}{n} * \frac{n(n+1)}{2}](https://tex.z-dn.net/?f=E%28x%29%20%3D%20%5Cfrac%7B1%7D%7Bn%7D%20%2A%20%5Cfrac%7Bn%28n%2B1%29%7D%7B2%7D)
![E(x) = \frac{n+1}{2}](https://tex.z-dn.net/?f=E%28x%29%20%3D%20%5Cfrac%7Bn%2B1%7D%7B2%7D)
Var(x) is calculated as:
![Var(x) = E(x^2) - (E(x))^2](https://tex.z-dn.net/?f=Var%28x%29%20%3D%20E%28x%5E2%29%20-%20%28E%28x%29%29%5E2)
Calculating: ![E(x^2)](https://tex.z-dn.net/?f=E%28x%5E2%29)
![E(x^2) = \sum \limits^{n}_{x=1} \ x^2 * \frac{1}{n}](https://tex.z-dn.net/?f=E%28x%5E2%29%20%3D%20%5Csum%20%5Climits%5E%7Bn%7D_%7Bx%3D1%7D%20%5C%20x%5E2%20%2A%20%5Cfrac%7B1%7D%7Bn%7D)
![E(x^2) = \frac{1}{n}\sum \limits^{n}_{x=1} \ x^2](https://tex.z-dn.net/?f=E%28x%5E2%29%20%3D%20%5Cfrac%7B1%7D%7Bn%7D%5Csum%20%5Climits%5E%7Bn%7D_%7Bx%3D1%7D%20%5C%20x%5E2)
Using the hint given:
![\sum \limits^{n}_{x=1} \ x^2 = \frac{n(n +1)(2n+1)}{6}](https://tex.z-dn.net/?f=%5Csum%20%5Climits%5E%7Bn%7D_%7Bx%3D1%7D%20%5C%20x%5E2%20%20%3D%20%5Cfrac%7Bn%28n%20%2B1%29%282n%2B1%29%7D%7B6%7D)
So:
![E(x^2) = \frac{1}{n} * \frac{n(n +1)(2n+1)}{6}](https://tex.z-dn.net/?f=E%28x%5E2%29%20%3D%20%5Cfrac%7B1%7D%7Bn%7D%20%2A%20%5Cfrac%7Bn%28n%20%2B1%29%282n%2B1%29%7D%7B6%7D)
![E(x^2) = \frac{(n +1)(2n+1)}{6}](https://tex.z-dn.net/?f=E%28x%5E2%29%20%3D%20%5Cfrac%7B%28n%20%2B1%29%282n%2B1%29%7D%7B6%7D)
So:
![Var(x) = E(x^2) - (E(x))^2](https://tex.z-dn.net/?f=Var%28x%29%20%3D%20E%28x%5E2%29%20-%20%28E%28x%29%29%5E2)
![Var(x) = \frac{(n+1)(2n+1)}{6} - (\frac{n+1}{2})^2](https://tex.z-dn.net/?f=Var%28x%29%20%3D%20%5Cfrac%7B%28n%2B1%29%282n%2B1%29%7D%7B6%7D%20-%20%28%5Cfrac%7Bn%2B1%7D%7B2%7D%29%5E2)
![Var(x) = \frac{(n+1)(2n+1)}{6} - \frac{n^2+2n+1}{4}](https://tex.z-dn.net/?f=Var%28x%29%20%3D%20%5Cfrac%7B%28n%2B1%29%282n%2B1%29%7D%7B6%7D%20-%20%5Cfrac%7Bn%5E2%2B2n%2B1%7D%7B4%7D)
![Var(x) = \frac{2n^2 +n+2n+1}{6} - \frac{n^2+2n+1}{4}](https://tex.z-dn.net/?f=Var%28x%29%20%3D%20%5Cfrac%7B2n%5E2%20%2Bn%2B2n%2B1%7D%7B6%7D%20-%20%5Cfrac%7Bn%5E2%2B2n%2B1%7D%7B4%7D)
![Var(x) = \frac{2n^2 +3n+1}{6} - \frac{n^2+2n+1}{4}](https://tex.z-dn.net/?f=Var%28x%29%20%3D%20%5Cfrac%7B2n%5E2%20%2B3n%2B1%7D%7B6%7D%20-%20%5Cfrac%7Bn%5E2%2B2n%2B1%7D%7B4%7D)
Take LCM
![Var(x) = \frac{4n^2 +6n+2 - 3n^2 - 6n - 3}{12}](https://tex.z-dn.net/?f=Var%28x%29%20%3D%20%5Cfrac%7B4n%5E2%20%2B6n%2B2%20-%203n%5E2%20-%206n%20-%203%7D%7B12%7D)
![Var(x) = \frac{4n^2 - 3n^2+6n- 6n +2 - 3}{12}](https://tex.z-dn.net/?f=Var%28x%29%20%3D%20%5Cfrac%7B4n%5E2%20-%203n%5E2%2B6n-%206n%20%2B2%20%20-%203%7D%7B12%7D)
![Var(x) = \frac{n^2 -1}{12}](https://tex.z-dn.net/?f=Var%28x%29%20%3D%20%5Cfrac%7Bn%5E2%20-1%7D%7B12%7D)
Apply difference of two squares
![Var(x) = \frac{(n+1)(n-1)}{12}](https://tex.z-dn.net/?f=Var%28x%29%20%3D%20%5Cfrac%7B%28n%2B1%29%28n-1%29%7D%7B12%7D)
Step-by-step explanation:
3(x-1) -5(x-4) = 8
3x-3 -5x+20 = 8
3x -5x -3 +20 = 8
-2x + 17 = 8
17 -8 = 2x
9 = 2x
2x = 9
x = 9/2
Answer:
dont cheat
Step-by-step explanation:
cheaters never win bro
Answer:
78
Step-by-step explanation:
when there is a five car diffrence its possible